1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

0.3 Boundary Value Problems 31
If the two ends of the rod are held at constant temperature, the boundary
conditions onuwould be

u( 0 )=T 0 , u(a)=T 1. (15)

On the other hand, if heat were supplied atx=0 (by a heating coil, for in-
stance), the boundary condition there would be

−κAdu

dx

( 0 )=H, (16)

whereHis measured in units of heat per unit time.

Example.
Solve the problem

−κ

d^2 u

dx^2 =−hu(x)

C

A,^0 <x<a, (17)

u( 0 )=T 0 , u(a)=T 0. (18)

(Physically, the rod is losing heat to a surrounding medium at temperature 0,
whilebothendsareheldatthesametemperatureT 0 .) If we designateμ^2 =
hC/κA, the differential equation becomes

d^2 u

dx^2 −μ

(^2) u= 0 , 0 <x<a,
with general solution
u(x)=c 1 cosh(μx)+c 2 sinh(μx).
Application of the boundary condition atx=0givesc 1 =T 0 ; the second
boundary condition requires that
u(a)=T 0 : T 0 =T 0 cosh(μa)+c 2 sinh(μa).
Thusc 2 =T 0 ( 1 −cosh(μa))/sinh(μa)and
u(x)=T 0

(

cosh(μx)+^1 −cosh(μa)

sinh(μa)

sinh(μx)

)

.

It should be clear now that solving a boundary value problem is not sub-
stantially different from solving an initial value problem. The procedure is (1)
find the general solution of the differential equation, which must contain some
arbitrary constants, and (2) apply the boundary conditions to determine val-
ues for the arbitrary constants. In our examples the differential equations have